Question

Graph $f\left(x\right)=x^2+4x-5$, by determining whether the graph opens up or down and by finding the vertex, axis of symmetry, and intercepts.

Short answer

The graph of the function is:

Step-by-step solution

Step 1. Given information.

The function to be plotted is:

$f\left(x\right)=x^2+4x-5$

The sign of $x^2$is positive, so graph will open upward.

Step 2. To find x-intercept and y-intercept of the function.

The function can be written as:

$y=x^2+4x-5$

$$\begin{gathered} y=x^2+2.2x+2^2-2^2-5 \\ y={(x+2)}^2-4-5 \\ y={\left(x+2\right)}^2-9 \end{gathered}$$

To find x-intercept:

$$\begin{gathered} {(x+2)}^2-9=0 \\ {(x+2)}^2-3^2=0 \\ (x+2-3)(x+2+3)=0 \\ (x-1)(x+5)=0 \\ x-1=0 \\ x=1 \\ x+5=0 \\ x=-5 \end{gathered}$$

$x-$intercepts are (1,0) and (-5,0)

To find y-intercept put $x=0$in the function.

localid="1647701094732" $$\begin{gathered} y={(0+2)}^2-9y=4-9 \\ y=-5 \end{gathered}$$

y-intercept is (0,-5)

Step 3. To find vertex and axis of symmetry of the graph.

The equation of the parabola:

$$\begin{gathered} {(x+2)}^2=(y+9) \\ (x-(-2\left)\right)=4.\frac{1}{4}(y-(-9\left)\right) \end{gathered}$$

By comparing the equation with standard form of parabola:${(x-h)}^2=4a(y-k)$, where $(h,k)$is vertex of the parabola.

We get $(h,k)=(-2,-9)$

Vertex is: (-2,-9)

The axis of symmetry is:

localid="1647700708833" $$\begin{gathered} x+2=0 \\ x=-2 \end{gathered}$$

Step 4. To draw the graph of the function.

The graph of the given function can be drawn as: